I am an undergraduate student working on a research project in theoretical chemistry, but I figure this particular obstacle is primarily a mathematical one. I will begin with a pretty long context, but the math problem itself is at the end. The following paper

is the main one on the topic that I will be asking about here (for anyone that may find it helpful), but I will provide the full context here.

Basically, there's this moving thing called a Gaussian wave packet that can be completely described with the parameters $x_t$ (center of the wave packet at time t), $p_t$ (momentum at time t), $\alpha_t$, (related to the width of the wavepacket), and $\gamma_t$ (related to the 'phase' of the wave packet. The first two parameters are real numbers, and the last two are complex. The subscript t indicates that these parameters are functions of time. Simply, my goal is to find the equations that explicitly state what these values are, given any value of time as an input. There are a LOT of big (by my standard), interconnected equations involved which make this a pretty big pain in the butt.

First of all, these Gaussian wave packets are meant to be related to a potential energy function, V. In our case,

$$V(x) = \frac{1}{2}m\omega^{2}x^{2} \tag{1}$$

This is the potential energy for a harmonic oscillator. m is a constant (mass), and $\omega$ is a constant (angular frequency). The energy of the system is given by:

$$H(p_t, x_t) = \frac{p_{t}^{2}}{2m} + V(x_t) \tag{2}$$


$$H(p_t, x_t) = \frac{p_{t}^{2}}{2m} + \frac{1}{2}m\omega^{2}x^{2} \tag{3}$$

Heller, the author of the paper provides equations of motion that should yield the golden answers of interest when solved:

$$\dot{x_t} = \frac{\partial H}{\partial p_t} \tag{4}$$

$$\dot{p_t} = -\frac{\partial H}{\partial x_t} \tag{5}$$

$$\dot{\alpha_t} = -(\frac{2}{m})\alpha_t^{2} - \frac{1}{2}V_{xx} \tag{6}$$

$$\dot{\gamma_t} = \frac{i\hbar\alpha_t}{m} + p_t\dot{x_t} - \frac{p_{t}^{2}}{2m} - \frac{1}{2}m\omega^{2}x^{2} \tag{7}$$


$$\dot{x_t} = \frac{dx_t}{dt} \tag{8}$$

and $V_{xx}$ is the second derivative of the potential with respect to x, at the value $x = x_t$.

Conveniently, Heller provides us with some solutions to this monstrosity in the event that $V(x) = \frac{1}{2}m\omega^{2}x^{2}$.

$$\alpha_t = -\frac{m\omega}{2}\Bigg(\frac{\frac{1}{2} m \omega - \alpha_0 \cot(\omega t)}{\alpha_0 + \frac{1}{2}m\omega \cot(\omega t)}\Bigg) \tag{9}$$

$$x_t = x_0 \cos(\omega t) + \frac{p_0}{m\omega}\sin(\omega t) \tag{10}$$

$$p_t = p_0 \cos(\omega t) - m\omega x_0 \sin(\omega t) \tag{11}$$

A subscript $0$ indicates that this is the initial value for that specific parameter.

Now, here is what kills me. Heller claims that if we set an initial value of $\alpha_0 = im\omega /2$, we get that $a_t = a_0$ for all values of time. That is, the $\alpha$ parameter becomes a fixed constant. And then, using this convenient starting point, he gives an answer for $\gamma$:

$$\gamma_t = -\frac{1}{4}i\hbar \ln(m\omega/\pi\hbar) - \frac{1}{2}\hbar\omega t + \frac{1}{2}(p_tx_t - p_0y_0) \tag{12}$$

This is very good if i want to have an initial value of $\alpha_0 = im\omega /2$, but I don't want to do that. I want to have a solution to $\gamma$ that is a function of $\alpha_0$, so that I can change my starting parameter for $\alpha$ to whatever I want and have my $\gamma$ value change accordingly.

What is the solution to equation (7)? Or at least, what can I do to solve it? I've tried using Wolfram and Symbolab to the best of my abilities, but they don't seem to come close in replicating something that resembles an answer. I have tried splitting up the problem into chunks and evaluating those separately, but again, these online calculators seem to give some very wacky results. If anybody could provide me with a solution (ideally also showing that they can derive eqn. (12) using their method by starting with $\alpha_0 = im\omega /2$ ), it would be a life saver for the progression of my research. However, any help towards an answer would be greatly appreciated.

  • 2
    $\begingroup$ What is (D)? Is it in the paper? $\endgroup$ Feb 11, 2019 at 1:08
  • $\begingroup$ hi, so sorry, just edited it. By (D) I meant eqn 7! $\endgroup$ Feb 11, 2019 at 4:14

1 Answer 1


I will provide a derivation of (12).

First, note that we have \begin{align} \dot \gamma_t=&\ \frac{i\hbar \alpha_t}{m}+\underbrace{p_t\dot x_t-H(p_t, x_t)}_{\text{Legendre Transform}}= \frac{i\hbar \alpha_t}{m}+L(\dot x_t, x_t) \end{align} where \begin{align} L(\dot x_t, x_t) = T-V= \frac{1}{2}m\dot x^2_t-\frac{1}{2}m\omega^2x_t^2 \end{align} is the Lagrangian of the $1$D harmonic oscillator.

Hence we have that \begin{align} \gamma_t =&\ \gamma_0+ \int^t_0 \frac{i\hbar \alpha_s}{m}+L(\dot x_s, x_s)\ ds=\ \gamma_0 -\frac{1}{2}\hbar \omega t+ \underbrace{\int^t_0 L(\dot x_s, x_s)\ ds}_{\text{Lagrangian action}}. \end{align}

As a standard exercise of integration by parts, you can show that \begin{align} \int^t_0 L(\dot x_s, x_s)\ ds = \frac{1}{2}(p_t x_t - p_0 x_0). \end{align}

Note the difficulties in the general setting are finding $\alpha_t$ and evaluating \begin{align} \frac{i\hbar}{m}\int^t_0\alpha_s\ ds. \end{align} But even that is just an exercise in the case of the $1$D harmonic oscillator since the equation for $\alpha_t$ is \begin{align} \dot \alpha_t=-\frac{2}{m}\alpha_t^2-m\omega^2 \end{align} which is just a separable equation.

  • $\begingroup$ Thanks a lot for the help, and sorry for a pretty major typo in the original post: I am looking for a solution to eqn 7. Your derivation of eqn 12 is very reassuring! Eqn 12 is a solution to eqn 7 when $\alpha_0 = im\omega/2$, but I am looking for a general solution that allows $\alpha_0$ to be any value. $\endgroup$ Feb 11, 2019 at 4:16
  • $\begingroup$ For the harmonic oscillator right? $\endgroup$ Feb 11, 2019 at 4:24
  • $\begingroup$ Please read the last paragraph carefully since I have already answered your concern. $\endgroup$ Feb 11, 2019 at 4:58
  • $\begingroup$ Thanks so much! Just to make sure I understand, the original post's eqn (9) gives a solution for $\alpha_t$ for any time t given an initial $\alpha_0$ value. If i slapped in some value of $\alpha_0$ such that the answer works out to be $\alpha_t = cos(t)$ (magically), my final gamma equation would be $\gamma_t = \gamma_0 + \frac{i\hbar}{m}[sin(t) - sin(0)] + \frac{1}{2}(p_tx_t - p_0x_0)$, correct? $\endgroup$ Feb 11, 2019 at 13:16
  • $\begingroup$ Assuming your expression for $\alpha_t$ is correct and you can find some initial data $\alpha_0$ to get cosine, which I doubt, then yes you are correct. $\endgroup$ Feb 11, 2019 at 13:46

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